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Question Number 45632 by maxmathsup by imad last updated on 14/Oct/18

$$1)find\int ln(1+x^3)dx$$$$2)calculate{\int }_0^{1}ln(1+x^3)ex$$

Commented by maxmathsup by imad last updated on 16/Oct/18

$$letI=\int ln(1+x^3)dxbyparts$$$$I=xln(1+x^3)-\int x\frac{3x^2}{1+x^3}dx=xln(1+x^3)-3\int \frac{1+x^3-1}{1+x^3}dx$$$$=xln(1+x^3)-3x+3\int \frac{dx}{1+x^3}letdecomposeF\left(x\right)=\frac{1}{1+x^3}$$$$F\left(x\right)=\frac{1}{(x+1)(x^2-x+1)}=\frac{a}{x+1}+\frac{bx+c}{x^2-x+1}$$$$a={lim}_{x\to -1}(x+1)F\left(x\right)=\frac{1}{3}$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=a+b\Rightarrow b=-a\Rightarrow F\left(x\right)=\frac{1}{3(x+1)}+\frac{-\frac{1}{3}x+c}{x^2-x+1}$$$$F\left(0\right)=1=\frac{1}{3}+c\Rightarrow c=\frac{2}{3}\Rightarrow F\left(x\right)=\frac{1}{3(x+1)}+\frac{-\frac{1}{3}x+\frac{2}{3}}{x^2-x+1}$$$$\Rightarrow 3F\left(x\right)=\frac{1}{x+1}+\frac{-x+2}{x^2-x+1}\Rightarrow \int 3F\left(x\right)dx=ln\mid x+1\mid -\frac{1}{2}\int \frac{2x-1-3}{x^2-x+1}$$$$=ln\mid x+1\mid -\frac{1}{2}ln\mid x^2-x+1\mid +\frac{3}{2}\int \frac{dx}{x^2-x+1}but$$$$\int \frac{dx}{x^2-x+1}=\int \frac{dx}{{(x-\frac{1}{2})}^2+\frac{3}{4}}{=}_{x-\frac{1}{2}=\frac{\sqrt{3}}{2}u}\frac{4}{3}\int \frac{1}{u^2+1}\frac{\sqrt{3}}{2}du$$$$=\frac{2}{\sqrt{3}}\int \frac{du}{1+u^2}=\frac{2}{\sqrt{3}}arctan\left(\frac{2x-1}{\sqrt{3}}\right)\Rightarrow \int 3F\left(x\right)dx$$$$=ln\mid x+1\mid -\frac{1}{2}ln(x^2-x+1)+\frac{3}{2}\frac{2}{\sqrt{3}}arctan\left(\frac{2x-1}{\sqrt{3}}\right)\Rightarrow$$$$\int \frac{dx}{1+x^3}=xln(1+x^3)-3x+ln\mid x+1\mid -\frac{1}{2}ln(x^2-x+1)+\sqrt{3}arctan\left(\frac{2x-1}{\sqrt{3}}\right)+c$$

Commented by maxmathsup by imad last updated on 16/Oct/18

$${\int }_0^1\frac{dx}{1+x^3}={[xln(1+x^3)-3x+ln\frac{\mid x+1\mid }{\sqrt{x^2-x+1}}+\sqrt{3}arctan\left(\frac{2x-1}{\sqrt{3}}\right)]}_0^1$$$$=ln\left(2\right)-3+ln\left(2\right)+\sqrt{3}arctan\left(\frac{1}{\sqrt{3}}\right)+\sqrt{3}arctan\left(\frac{1}{\sqrt{3}}\right)$$$$=2ln\left(2\right)-3+2\sqrt{3}\frac{\pi }{6}=2ln\left(2\right)+\frac{\pi \sqrt{3}}{3}-3.$$

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